package com.cg.offer;

import org.junit.Test;

import java.util.Deque;
import java.util.LinkedList;
import java.util.Queue;

/**
 * @program: LeetCode->Offer_59_II
 * @description: 剑指Offer 59-II.队列的最大值
 * @author: cg
 * @create: 2022-04-04 17:22
 **/
public class Offer_59_II {

    @Test
    public void test59_II() {
        MaxQueue maxQueue = new MaxQueue();
        maxQueue.push_back(1);
        maxQueue.push_back(2);
        System.out.println(maxQueue.max_value());
        System.out.println(maxQueue.pop_front());
        System.out.println(maxQueue.max_value());
    }

}

/**
 * 请定义一个队列并实现函数 max_value 得到队列里的最大值，要求函数max_value、push_back 和 pop_front 的均摊时间复杂度都是O(1)。
 * <p>
 * 若队列为空，pop_front 和 max_value 需要返回 -1
 * <p>
 * 示例 1：
 * 输入:
 * ["MaxQueue","push_back","push_back","max_value","pop_front","max_value"]
 * [[],[1],[2],[],[],[]]
 * 输出: [null,null,null,2,1,2]
 * <p>
 * 示例 2：
 * 输入:
 * ["MaxQueue","pop_front","max_value"]
 * [[],[],[]]
 * 输出: [null,-1,-1]
 * <p>
 * 限制：
 * 1 <= push_back,pop_front,max_value的总操作数 <= 10000
 * 1 <= value <= 10^5
 */
class MaxQueue {
    Queue<Integer> queue;
    Deque<Integer> deque;

    public MaxQueue() {
        queue = new LinkedList<>();
        deque = new LinkedList<>();
    }

    public int max_value() {
        return deque.isEmpty() ? -1 : deque.peekFirst();
    }

    public void push_back(int value) {
        queue.offer(value);
        while (!deque.isEmpty() && deque.peekLast() < value) {
            deque.pollLast();
        }
        deque.offerLast(value);
    }

    public int pop_front() {
        if (queue.isEmpty()) {
            return -1;
        }
        if (queue.peek().equals(deque.peekFirst())) {
            deque.pollFirst();
        }
        return queue.poll();
    }
}